Calculus III, Mathematics 2210-90
Examination 2, October 16,18, 2003
You may use graphing calculators and a Table of Integrals. Each problem is worth 20 points. You MUST show your work. Just the correct answer is not sufficient for any points .
1a. Write down an equation of a hyperbola whose major axis is the line y = x .
Solution .
xy = 1 will do. So will any form of the type Ax 2
B 2 > 4 A 2 . Another idea would be to start with u 2 − v 2
+ Bxy + Ay 2 = D , with
= 1, and make the substitution u = x + y, v = x − y , which leads to 4 xy = 1.
1b. Write down an equation of an ellipse (not a circle) whose major axis is the line y = x .
Solution . Any form of the type Ax 2 idea would be to start with 4 u 2 + v 2
+ Bxy + Ay 2 = D , with B 2 < 4 A 2 will do. Another
= 1, and make the substitution u = x + y, v = x − y , which leads to 5 x 2 + 6 xy + 5 y 2 = 1.
2. Let f ( x, y ) = x 2 + 3 xy + 4 y 2 .
a) What is ∇ f ?
∇ f = (2 x + 3 y ) I + (3 x + 8 y ) J .
b) Find the equation of the line tangent to the curve f ( x, y ) = 14 at the point (2,1).
Solution . At X
0
: (2 , 1) , ∇ f = 7 I + 14 J . The equation of the line is which is 7( x − 2) + 14( y − 1) = 0, which simplifies to x + 2 y = 4.
∇ f · ( X − X
0
) = 0,
3. Let w = x 2 − yz . Let X ( t ) = e t I + e t +1 for dw/dt along Γ as a function of t .
J + e 2 t K describe the curve Γ. Find the formula
Solution .
∇ w = 2 x I − z J − y K , so, along Γ, in terms of t , ∇ w = 2 e t I − e 2 t J − e t +1 K .
Now d X dt
= e t I + e t +1 J + 2 e 2 t K , and dw dt
= ∇ w · d X dt
= 2 e 2 t − e 3 t +1 − 2 e 3 t +1 = 2 e 2 t − 3 e 3 t +1 .
4. Let f ( x, y ) = x 2 y + 27 y 2 + x .
Solution . a) ∇ f = (2 xy + 1) I + ( x 2 + 54 y ) J b) What are the critical points of f ?
1

Solution . We must solve the pair of equations 2 xy + 1 = 0 , x 2 y = − 1 / 2 x ; substituting in the second gives x 3 = 27, so x
+ 54
= 3 and y y
= 0. The first gives
= − 1 / 6. The only critical point is (3,-12/6).
c) What kind of critical points are they?
Solution . We calculate f xx the point is a saddle point.
= − 1 / 3 , f xy
= 6 , f yy
= 54, so f xx f yy
− f 2 xy
= − 54 < 0, and
5. Find the minimum value of 3 x 2 + y 2 on the curve x 2 + xy = 1.
Solution . We use the method of Lagrange multipliers. Let f ( x, y ) = 3 x 2 x 2 + xy . We have
∇ f = 6 x I + 2 y J , ∇ g = (2 x + y ) I + x J .
+ y 2 , g ( x, y ) =
The equations to solve are
3 x = λ (2 x + y ) , y = λx , x 2 + xy = 1 .
Substitute the second equation in the first to get 3 x = λ (2 x + λx ). Since x = 0 does not solve the last equation, we can cancel x to get 3 = 2 λ + λ 2 , which has the solutions
λ = 1 , − 3. The case λ = 1 gives us y = x ; substitute that in the last equation to get
2 x 2 = 1, or x = y = ± 1 /
√
2, and f ( x, y ) = 2. The case λ = − 3 leads to y = − 3 x ; putting that in the last equation gives − x 2 = 1, an impossibility. Thus the only possible solution is at the first point. Since f is always positive, it must have a minimum, so since 2 is the only candidate, it is the minimum.
2